YES
Termination Proof
Termination Proof
by ttt2 (version ttt2 1.15)
Input
The rewrite relation of the following TRS is considered.
|
a(a(x0)) |
→ |
c(b(x0)) |
|
b(b(x0)) |
→ |
c(a(x0)) |
|
c(c(x0)) |
→ |
b(a(x0)) |
Proof
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
a(a(x0)) |
→ |
b(c(x0)) |
|
b(b(x0)) |
→ |
a(c(x0)) |
|
c(c(x0)) |
→ |
a(b(x0)) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a#(a(x0)) |
→ |
c#(x0) |
|
a#(a(x0)) |
→ |
b#(c(x0)) |
|
b#(b(x0)) |
→ |
c#(x0) |
|
b#(b(x0)) |
→ |
a#(c(x0)) |
|
c#(c(x0)) |
→ |
b#(x0) |
|
c#(c(x0)) |
→ |
a#(b(x0)) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
| [c#(x1)] |
= |
1 ·
x1 +
-∞
|
| [b(x1)] |
= |
1 ·
x1 +
-∞
|
| [a#(x1)] |
= |
0 ·
x1 +
-∞
|
| [b#(x1)] |
= |
0 ·
x1 +
-∞
|
| [a(x1)] |
= |
1 ·
x1 +
-∞
|
| [c(x1)] |
= |
1 ·
x1 +
-∞
|
together with the usable
rules
|
a(a(x0)) |
→ |
b(c(x0)) |
|
b(b(x0)) |
→ |
a(c(x0)) |
|
c(c(x0)) |
→ |
a(b(x0)) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
a#(a(x0)) |
→ |
c#(x0) |
|
a#(a(x0)) |
→ |
b#(c(x0)) |
|
b#(b(x0)) |
→ |
c#(x0) |
|
b#(b(x0)) |
→ |
a#(c(x0)) |
remain.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.